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Choose the convention ${\psi }^{\prime }=g\psi$ for a field in the defining representation and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. Requiring $D_{\mu }^{\prime }{\psi }^{\prime }=g{D}_{\mu }\psi$ gives the Yang-Mills gauge transformationThe gauge field strength is the curvature $[D_{\mu },D_{\nu }]=F_{\mu \nu }$, soFor an adjoint field $Y$, the adjoint covariant derivative is $D_{\mu }Y={\partial }_{\mu }Y+[A_{\mu },Y]$, and direct substitution gives $D_{\mu }^{\prime }{Y}^{\prime }=g(D_{\mu }Y)g^{-1}$. Consequently $D_{\mu }{F}_{\nu \rho }$ transforms by conjugation too.

The Killing-form invariance under a Lie-group adjoint action gives $\kappa (gY{g}^{-1},gZ{g}^{-1})=\kappa (Y,Z)$. Its proof is that the adjoint matrices are conjugated by the same invertible map, leaving their trace product unchanged. With $s=\kappa (F_{\mu \nu },F^{\mu \nu })$, both $s^p$ and $\kappa (D_{\mu }{F}_{\nu \rho },D^{\mu }{F}^{\nu \rho })$ are therefore gauge invariant.

For their field equations, take a compactly supported variation $a_{\mu }=\delta A_{\mu }$. Curvature varies byThe invariant form permits gauge-covariant integration by parts, because ${\partial }_{\mu }\kappa (Y,Z)=\kappa (D_{\mu }Y,Z)+\kappa (Y,D_{\mu }Z)$. Antisymmetry of $F$ givesIntegrating, the power of the Yang-Mills invariant gives the equationFor $p=1$ this recovers the Yang-Mills equations. For larger $p$, keep the displayed product form, including at $s=0$.

For the derivative-squared action, write $B_{\mu \nu \rho }=D_{\mu }{F}_{\nu \rho }$. Varying the derivative as well as the curvature givesIts first variation isIntegrate the first term once, substitute $\delta F_{\nu \rho }=D_{\nu }{a}_{\rho }-D_{\rho }{a}_{\nu }$, and integrate again. It becomes $4\int \kappa (a_{\nu },D_{\mu }{D}^2{F}^{\mu \nu })d^{4}x$, where $D^2=D_{\lambda }{D}^{\lambda }$. Invariance converts the second term into $2\int \kappa (a_{\nu },[F_{\rho \sigma },D^{\nu }{F}^{\rho \sigma }])d^{4}x$. Nondegeneracy of the Killing form and arbitrary variations yield the derivative-squared Yang-Mills action equationThe derivative order matters because covariant derivatives generally do not commute. In an abelian limit the commutator term vanishes and the equation reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$, a useful check on the fourth-order equation for the potential.

The Noether gauging procedure starts by allowing a constant symmetry parameter to depend on position. For an internal symmetry written as $\delta {\varphi }^i={ϵ}^A{T}_A{\varphi }^i$, integration by parts identifies the Noether current:up to a boundary term. Introduce a gauge field $A_{\mu }^A$ and add $-g{A}_{\mu }^A{j}_A^{\mu }$, with the inhomogeneous gauge transformation $\delta A_{\mu }^A=g^{-1}{\partial }_{\mu }{ϵ}^A+\cdots$. This cancels the displayed variation. The current coupling has its own variation, so one iterates: add the needed higher-order interactions and corrections to the gauge transformations, and require closure of the local symmetry algebra. For an ordinary gauge theory, this reorganizes derivatives into gauge covariant derivatives and fixes the nonlinear gauge field strength. Gauging is therefore more than replacing a constant parameter by a function.

For rigid supersymmetry, the corresponding Noether current is the spinor-valued supercurrent $S^{\mu }$. A local Majorana spinor parameter gives a variation proportional to $({\partial }_{\mu }\overline{ϵ})S^{\mu }$. A gravitino ${\Psi }_{\mu }$, with an inhomogeneous transformation $\delta {\Psi }_{\mu }\sim {\kappa }^{-1}{\partial }_{\mu }ϵ$, cancels it through a coupling proportional to $\kappa {\bar{\Psi }}_{\mu }{S}^{\mu }$. The commutator of two rigid supersymmetry transformations is a translation. With local parameters, the translation parameter becomes position dependent, so the completion must also gauge translations: it introduces the vierbein, diffeomorphism invariance of general relativity, and local Lorentz transformations. The gravitino and graviton thus belong to the same supergravity multiplet.

For pure minimal four-dimensional supergravity, start with the free massless Fierz-Pauli action and the free Rarita-Schwinger field kinetic term. The Noether gauging procedure replaces the former by the Einstein-Hilbert action and couples the latter to the vierbein and spin connection. In a canonical normalization the leading transformations can be writtenAn overall sign or phase changes with the curvature and Majorana spinor conventions, but the relative normalization must be used consistently. To see why these terms fit together, first work to quadratic order in fermions, so the spin connection is torsion free. Varying the Rarita-Schwinger field kinetic term under $\delta {\Psi }_{\mu }={\nabla }_{\mu }ϵ/\kappa$ produces an antisymmetric double spinor covariant derivative. The spinor curvature identity and the Einstein tensor contraction of spin curvature give, in the matching curvature convention,The resulting Einstein tensor term cancels the variation of the Einstein-Hilbert action under $\delta e_{\mu }{}^a$. This is the essential first nontrivial cancellation in the construction.

At higher fermion order, the variations of the vierbein and spin connection produce further terms. Treating the spin connection as independent makes its Euler-Lagrange field equation algebraic. Its solution is a gravitino-induced torsion connection. With ${\Psi }_a=e_a{}^{\mu }{\Psi }_{\mu }$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, a conventional canonical normalization givesEquivalently, with $T^a=d{e}^a+{\tilde{\omega }}^a{}_b\wedge e^b$, the torsion tensor obeysThe Majorana Grassmann bilinear interchange makes the right-hand side antisymmetric in $\mu ,\nu$. Eliminating the spin connection generates the four-fermion completion. In the 1.5-order formalism one substitutes its solution but varies the other fields holding the connection fixed: the omitted chain-rule term is $(\delta S/\delta \omega )\delta \tilde{\omega }=0$. The remaining cubic fermion variations cancel by the Clifford algebra and Fierz rearrangements. This produces the displayed supergravity action in its compact connection-dependent form. Calling this formulation on shell means that no independent auxiliary fields have been retained and that closure of the supersymmetry algebra uses the propagating Euler-Lagrange field equations; it does not mean that action invariance is obtained only by setting every variation to zero on a solution.

In the antisymmetrized kinetic term, ${\Psi }_{\sigma }$ is a spinor-valued one-form and the derivative is conventionally writtenThis is the Lorentz-covariant exterior-derivative convention for that term. If one instead uses the full affine covariant derivative of a vector-spinor, a coordinate-index connection is also present and the antisymmetrization must include its torsion tensor correction. The crucial physical distinction from pure metric general relativity is the spin connection contains gravitino-induced contorsion, rather than being the torsion-free Levi-Civita connection. It reduces to the ordinary spinor covariant derivative when the gravitino vanishes.

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.

The commutator of two gauge covariant derivatives defines the gauge field strength:ThusConjugating the derivative commutator gives $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$, so the gauge field strength transforms homogeneously in the Adjoint representation:Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.

Only $g_{(ab)}$ contributes to the Yang-Mills theory Lagrangian, since $F_{\mu \nu }^a{F}^{b\mu \nu }$ is symmetric in $a,b$. Thus take $g_{ab}$ to be a constant real symmetric bilinear form. Its infinitesimal variation isArbitrary local field strengths and gauge parameters give the necessary and sufficient invariant gauge kinetic form conditionAn antisymmetric part is unconstrained but contributes nothing. A nondegenerate kinetic term additionally needs a nondegenerate bilinear form; positive energy additionally needs a positive internal metric in the signature $(+---)$ convention. Neither extra condition follows solely from gauge invariance. For a compact simple real Lie algebra, $-\kappa$ is positive definite and is the standard choice up to a positive factor.

For the intended complex-simple setting, the Killing form is nondegenerate. Consequently write any invariant bilinear form on a Lie algebra uniquely as $g(X,Y)=\kappa (SX,Y)$. Invariance of both forms impliesfor every $X,Y,Z$, so $S$ commutes with the Adjoint representation. A subspace invariant under all adjoint maps is an ideal, hence simplicity makes the adjoint representation irreducible. By Schur lemma, $S=cI$, giving the uniqueness of an invariant bilinear form on a simple Lie algebra:The same conclusion holds for compact simple real algebras by complexification, with $c$ real; nondegeneracy excludes $c=0$, and positive energy selects $c<0$ in the stated convention.

There is a genuine real-simple exception to uniqueness of the Killing form if “simple” is read as an arbitrary real simple algebra. Regard ${\mathfrak{sl}}_2(\mathbb{C})$ as real. It is real simple: its complexification consists of two complex simple factors exchanged by conjugation, so a conjugation-stable ideal is zero or the whole algebra. Both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ are real symmetric invariant forms, while ${\kappa }_{\mathbb{R}}=2\mathrm{Re}{\kappa }_{\mathbb{C}}$. With $H=\mathrm{diag}(1,-1)$, the Killing form of the special linear Lie algebra gives ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$. Therefore the imaginary part cannot be a real multiple of ${\kappa }_{\mathbb{R}}$. The uniqueness conclusion requires the customary complex-simple or compact-real interpretation.

Take a smooth gauge transformation $g:{\mathbb{R}}^3\to SU(2)$. In the anti-Hermitian convention of the adjoint covariant derivative, setExpanding ${\partial }_j(g\Phi g^{-1})$ shows that its two differentiated-$g$ terms cancel with the extra terms in $[A_j^g,{\Phi }^g]$. Thus $D_j^g{\Phi }^g=g(D_j\Phi )g^{-1}$. The gauge field strength is $F_{jk}={\partial }_j{A}_k-{\partial }_k{A}_j+[A_j,A_k]$, and the same cancellation, or the commutator of covariant derivatives, gives $F_{jk}^g=g{F}_{jk}{g}^{-1}$.

The matrix trace is unchanged by conjugation, soEach field-strength norm, covariant-gradient norm and Higgs norm is therefore invariant pointwise. It follows that $V_{\lambda }(A^g,{\Phi }^g)=V_{\lambda }(A,\Phi )$. The displayed tensor made from these fields is also gauge invariant wherever $\Phi \ne 0$.

There is a normalization problem in the requested final identity. With the ordinary defining $2\times 2$ trace and the printed inner product, an explicit orthonormal basis isThe last identity follows from the Pauli matrix multiplication law. In particular it is not permissible to combine this orthonormality with unit structure constants of a Lie algebra.

Write $A_j=a_j^a{e}_a$ and $\Phi =\phi e_3$ with $\phi \ne 0$, and put $s=\mathrm{sgn}\phi$. Define $C_{jk}=a_j^1{a}_k^2-a_j^2{a}_k^1$. ThenOnly the bracket of the transverse terms contributes in the $e_3$ direction, givingHence the literal tensor printed in the question isThe transverse term does not cancel with this trace normalization of su(2).

A concrete counterexample to the printed closure assertion is $\Phi =e_3$, $A_1=e_1$, $A_2=x_3{e}_2$, $A_3=0$ on a ball. Directly, $F_{12}=\sqrt{2}{x}_3{e}_3$, $F_{23}=-e_2$ and $F_{31}=0$, while $D_1\Phi =-\sqrt{2}{e}_2$, $D_2\Phi =\sqrt{2}{x}_3{e}_1$ and $D_3\Phi =0$. It follows thatThe Higgs field never vanishes. If global finite energy is desired, multiply both nonzero gauge-potential components by a smooth compactly supported cutoff equal to one on this ball and keep $\Phi =e_3$ everywhere. The energy is finite and the same local counterexample remains. Thus the claimed closure is false with the ordinary trace normalization actually printed; it is not a consequence of gauge invariance alone.

The consistent 't Hooft electromagnetic tensor for the printed inner product is insteadIn the directional gauge, its commutator coefficient is $1/{\kappa }^2$, so the terms cancel exactly andOn a connected region where $\phi \ne 0$, $s$ is constant. Commutation of ordinary partial derivatives provesAt any nonzero-Higgs point, use the permitted local gauge transformation to align the field with a fixed internal direction. Since $\mathcal{F}$ is gauge invariant, the identity derived in that gauge holds in the original gauge on the same ball. These balls cover every nonzero-Higgs region, proving that the corrected tensor is a closed differential form there.

Alternatively, keeping the tensor's printed coefficient one but replacing the inner product by $-2\mathrm{tr}(XY)$ gives the orthonormal basis $e_a=-i{\sigma }_a/2$ with $\kappa =1$. The same calculation then gives $f_{jk}=s({\partial }_j{a}_k^3-{\partial }_k{a}_j^3)$ and the intended identity. Either repair makes the bracket and inner-product normalizations agree; they must not be silently mixed.

Use matrix multiplication together with the wedge product of differential forms, suppressing wedge symbols in this calculation. Set $F=dA+A^2$ and $a=\delta A$. The gauge field strength varies as $\delta F=da+Aa+aA$, the covariant exterior derivative of $a$. The matrix trace has graded cyclicity, $\mathrm{Tr}(\alpha \beta )=(-1{)}^{pq}\mathrm{Tr}(\beta \alpha )$ for forms of degrees $p,q$. Thus the Chern-Simons 5-form can be expanded asFor example, the two cross terms in $F^{2}A$ both become $\mathrm{Tr}(A^{3}dA)$; combining the remaining two terms gives the displayed coefficients. This expansion keeps track of the noncommuting products rather than treating $A$ as a scalar one-form.

Put $B=dA$. Vary the first term and use the graded exterior derivative to move derivatives off $a$:Here $dB=0$. For the next term use $d(A^3)=B{A}^2-ABA+A^{2}B$. Graded cyclicity then givesFinally all five contributions to $\delta \mathrm{Tr}(A^5)$ are equal under the matrix trace, giving $5\mathrm{Tr}(a{A}^4)$. Combining these results proves the variation of the Chern-Simons 5-form:Consequently, on an oriented five-manifold $N$,The boundary term vanishes for a closed manifold, for variations supported in its interior, or when the tangential boundary connection is held fixed. A different boundary variational problem would require boundary conditions or a boundary action as well.

Write $A=A^{\alpha }{T}_{\alpha }$ in a basis of the Lie algebra. Independent compactly supported variations of each component yield the bulk Euler-Lagrange equationsEquivalently, with $d_{\alpha \beta \gamma }=\frac{1}{2}\mathrm{Tr}[T_{\alpha }\{T_{\beta },T_{\gamma }\}]$, these are $d_{\alpha \beta \gamma }{F}^{\beta }\wedge F^{\gamma }=0$. The symmetry of the two curvature two-forms accounts for the anticommutator. Writing simply $F\wedge F=0$ as a full matrix equation requires an additional assumption: variations must span a matrix algebra paired nondegenerately by the matrix trace. For a smaller Lie algebra, only its trace-paired projection is constrained. If the trace is literally taken in the adjoint representation of a semisimple Lie algebra, its symmetric cubic invariant vanishes, and these bulk equations are identically zero. Indeed, the adjoint generators are skew with respect to the nondegenerate invariant Killing form; transposition in this pairing reverses the sign of the symmetrized trace of three generators. The usual nontrivial five-dimensional theory therefore needs an invariant cubic trace that does not vanish; saying that the gauge field transforms in the adjoint alone does not specify such a trace.

Under a Yang-Mills gauge transformation, use $A^g=g^{-1}Ag+g^{-1}dg$, so $F^g=g^{-1}Fg$. For any Lie-algebra element $T$,Since $gT{g}^{-1}$ is still in the Lie algebra, the right-hand side vanishes whenever the original field satisfies every equation. Every gauge transform of a bulk solution remains a bulk solution. This conclusion follows from covariance of the field equations; it does not require the action itself to be exactly invariant under all large gauge transformations or under transformations changing prescribed boundary data.

Use the mostly-plus Clifford algebra and Riemann curvature tensor conventions of the preceding solutions, with $e=\det (e_{\mu }{}^a)$, ${\kappa }^2=8\pi G$, and a Majorana gravitino. An action for minimal four-dimensional supergravity in the exact transformation normalization given isThe kinetic coefficient is tied to the normalization of the supersymmetry transformations: if ${\Psi }_{\mu }=2{\psi }_{\mu }$ and $\eta =2ϵ$, this becomes the canonical coefficient $-1/2$, with $\delta e_{\mu }{}^a=(\kappa /2)\bar{\eta }{\gamma }^a{\Psi }_{\mu }$ and $\delta {\Psi }_{\mu }={\nabla }_{\mu }\eta /\kappa$. Mixing those two normalizations would spoil the cancellation.

To the retained order, use the torsion-free connection. More generally the 1.5-order formalism allows its variation to be omitted after imposing its own algebraic equation; the resulting contorsion is quadratic in fermions and first affects the displayed action at quartic order. The metric tensor variation induced by the vierbein is $\delta g_{\mu \nu }=4\kappa \overline{ϵ}{\gamma }_{(\mu }{\psi }_{\nu )}$, so the Einstein-Hilbert action varies asFor the fermionic action, varying both Majorana factors gives the same contribution after integration by parts and the Grassmann bilinear interchange rule. ThusThe vector-index Riemann curvature tensor contribution vanishes by the first Bianchi identity. The spin-index contribution is determined byFor the last equality, expand the antisymmetric three-gamma/two-gamma product: the totally antisymmetric Riemann curvature tensor terms vanish by Bianchi, and the surviving terms are the Ricci contraction minus half its trace. Hencewhich cancels the gravitational variation. Variations of vierbeins, gamma matrices and the connection in the fermionic kinetic term carry three dynamical fermions. They lie beyond this linear-fermion cancellation and are cancelled by the omitted four-fermion completion. Thus the action is invariant up to boundary terms at the order specified, rather than claiming that a truncated action is exactly invariant at every fermion order.

For the cosmological extension, add a gravitino mass parameter $m$ and deform the derivative:Indeed the fermion terms combine as $-2{\bar{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\mathcal{D}}_{\nu }{\psi }_{\rho }$ because ${\gamma }^{\mu \nu \rho }{\gamma }_{\nu }=-2{\gamma }^{\mu \rho }$. The modified spinor connection curvature isand contraction givesThe same action-variation cancellation therefore works with $G$ replaced by $G+\Lambda g$ precisely forThis is the Anti-de Sitter deformation of minimal supergravity. A cosmological term alone would leave its variation uncancelled; the mass term and the gamma shift of the gravitino transformation are both necessary. A positive cosmological constant cannot arise from this real unbroken minimal deformation.

For electromagnetism, retain $\mathcal{N}=1$ and introduce an Abelian vector multiplet $(A_{\mu },\lambda )$, with Majorana gaugino $\lambda$. A complete quadratic-fermion extension isThe accompanying leading transformations areThey preserve the rigid Maxwell-gaugino action for constant $ϵ$. For a local parameter, the variation containsTo obtain this, vary $F^2$, vary both gaugino factors, and use ${\gamma }^{ab}{\gamma }^{\mu }={\gamma }^{ab\mu }+g^{b\mu }{\gamma }^a-g^{a\mu }{\gamma }^b$. The terms with $\nabla \lambda$ cancel between the two actions; integrating the three-gamma term uses ${\nabla }_{[\mu }{F}_{\nu \rho ]}=0$, leaving exactly the displayed derivative of the parameter. The interaction is $-\kappa e{\bar{\psi }}_{\mu }{j}^{\mu }$, whose variation under $\delta {\bar{\psi }}_{\mu }={\nabla }_{\mu }\overline{ϵ}/\kappa$ cancels that term.

The other linear-fermion variation is also fixed. Put $\mathcal{F}=F_{ab}{\gamma }^{ab}$ and $T_{\mathrm{EM}}^{\mu \nu }=F^{\mu \rho }{F}^{\nu }{}_{\rho }-g^{\mu \nu }{F}^2/4$. Clifford multiplication givesVarying $\lambda$ in the gravitino-current interaction then gives $2\kappa e{T}_{\mathrm{EM}}^{\mu \nu }{\bar{\psi }}_{\mu }{\gamma }_{\nu }ϵ=-2\kappa e{T}_{\mathrm{EM}}^{\mu \nu }\overline{ϵ}{\gamma }_{\nu }{\psi }_{\mu }$, which cancels the Maxwell vierbein variation. This proves the required supergravity coupling of an Abelian vector multiplet through the retained order.

Beyond that order one uses a supercovariant gauge field strength, here ${\hat{F}}_{\mu \nu }=F_{\mu \nu }-2\kappa {\bar{\psi }}_{[\mu }{\gamma }_{\nu ]}\lambda$, and the associated higher-fermion transformation terms and four-fermion interactions. An auxiliary scalar may be retained for off-shell closure, or eliminated in this on-shell formulation. A Maxwell field without a fermionic partner cannot be coupled by merely adding $-F^2/4$ while leaving all transformations unchanged. An alternative is an $\mathcal{N}=2$ gravity multiplet, where the vector is a graviphoton and a second gravitino supplies its extra fermionic states; the explicit construction above is the minimal $\mathcal{N}=1$ matter extension.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.